The Maclaurin series for sin x is use the corresponding Maclaurin polynomial of degree 5 to approximate integration from -1 to 1 i dont have any idea to solve it..
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Originally Posted by nameck The Maclaurin series for sin x is use the corresponding Maclaurin polynomial of degree 5 to approximate integration from -1 to 1 i dont have any idea to solve it.. Integrate "term by term"! Since Since the problem said "polynomial of degree 5" and integrating will give degree 5, you can stop there.
got it!! =)
The McLaurin expasion of the sin function is... (1) Only the 'odd' powers of x give contribution, so that is... (2) ... no matter which is a... Merry Christmas from Italy
Yes, but this was sin(x)/x which is an even function. Originally Posted by chisigma The McLaurin expasion of the sin function is... (1) Only the 'odd' powers of x give contribution, so that is... (2) ... no matter which is a... Merry Christmas from Italy
I have posted that in order to clarify the misundestanding caused by the fact that the function to be integrated is and not as written is the first post by nameck: the Maclaurin series for sin x is... Merry Christmas from Italy
i got 1.8922222.. is my answer correct?
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